Here, u_x = 72 km h^(-1) = 72 xx 5/ (18) = 20 ms^(-1)`
`y= 2.0 m`
taking verticl downward motion of the stone from the window of bus upto road, we bave
u_y =0. y= 2 . M , a_y = 10 ms^(-2) , t= ?`
As ` y=u_y t = 1/2 a_y t^2`
:. ` 2.0 =0 + 1/2 xx 10 xx t^2 = 5 t^2`
`t^2 = ( 2.0) /5 = 0.4 ` or t= sqrt . 0 4 s`
Taking horizontal motion of the stone from the window of bus upto point of striking the road , we have
` u_x = 20 ms^(1) , a_y =0 . t = sqrt 0.4 s, x= ?`
` x= u_x t + 1/2 a_x t^2`
` =20 xx sqrt 0.4 + 1/2 xx 0 xx ( 0.4) = 20 sqrt 0.4`
` = 4 sqrt 10 = 4 xx 3. 162 = 12 . 65 m`.